A mixture of carbon and sulfur has a mass of 9.0 g. Complete combustion with excess O_2 gives 27.9g of a mixture of CO_2 and SO_2.?

Relative atomic masses : C = 12.01, S = 32.06
Relative molecular masses : CO₂ = 44.01, SO₂ = 64.06

Let a g and b g be the mass of C and that of S in the mixture respectively.

No. of moles of C reacted = (a g) / (12.01 g/mol) = a/12.01 mol
No. of moles of CO₂ formed = a/12.01 mol
Mass of CO₂ formed = (a/12.01 mol) × (44.01 g/mol) = 44.01a/12.01 g = 3.664a g

S + O₂ → SO₂
No. of moles of S reacted = (b g) / (32.06 g/mol) = b/32.06 mol
No. of moles of CO₂ formed = b/32.06 mol
Mass of CO₂ formed = (b/32.06 mol) × (64.06 g/mol) = 64.06b/32.06 g = 1.998b g

Total mass of the mixture of C and S (in g) :
a + b = 9.0 ....... [1]

Total mass of the mixture of CO₂ and SO₂ (in g) :
3.664a + 1.998b = 27.9 ...... [2]

[1] * 3.664 :
3.664a + 3.664b = 32.98 ...... [3]

[3] - [2] :
1.666b = 5.08
b = 3.05

Mass of S in the mixture = 3.05 g

x= mass Carbon ...
x + S = 9 .... Cmass S = 9 - x
9 + 18.9 = 27.9 ... so mass O2 combined with both = 18.9


C + O2 --> CO2 ----- 12gC/32gO ... xgC / y gO
(32/12) x = 8x/3

S + O2 --> SO2 --- 32gS/32gO ... (9 - x)g S / z g O
z g O = (9 - x)

total O = 18.9 = (9 - x) + 8x/3
56.7 = 27 - 3x + 8x
5x = 29.9
x = 5.98 grams C
9 - x = 3.02 g S